Finite Element Analysis. Barna Szabó

Finite Element Analysis

u comma v right-parenthesis minus upper F left-parenthesis v right-parenthesis equals 0"/> in the following form:

(1.45) StartSet b EndSet Superscript upper T Baseline left-parenthesis left-parenthesis left-bracket upper K right-bracket plus left-bracket upper M right-bracket right-parenthesis StartSet a EndSet minus StartSet r EndSet right-parenthesis equals 0 period

Since this must hold for any choice of StartSet b EndSet , it follows that

(1.46) left-parenthesis left-bracket upper K right-bracket plus left-bracket upper M right-bracket right-parenthesis StartSet a EndSet equals StartSet r EndSet

which is the same system of linear equations we needed to solve when minimizing the integral script upper I , see eq. (1.14). Of course, this is not a coincidence. The solution of the generalized problem: “Find u Subscript n Baseline element-of upper S such that upper B left-parenthesis u Subscript n Baseline comma v right-parenthesis equals upper F left-parenthesis v right-parenthesis for all v element-of upper V ”, minimizes the error in the energy norm. See Theorem 1.4.

Theorem 1.3 The error e defined by e equals u minus u Subscript n satisfies upper B left-parenthesis e comma v right-parenthesis equals 0 for all v element-of upper S Superscript 0 Baseline left-parenthesis upper I right-parenthesis . This result follows directly from

StartLayout 1st Row 1st Column upper B left-parenthesis u comma v right-parenthesis equals 2nd Column upper F left-parenthesis v right-parenthesis for all v element-of upper S Superscript 0 Baseline left-parenthesis upper I right-parenthesis 2nd Row 1st Column upper B left-parenthesis u Subscript n Baseline comma v right-parenthesis equals 2nd Column upper F left-parenthesis v right-parenthesis for all v element-of upper S Superscript 0 Baseline left-parenthesis upper I right-parenthesis period EndLayout

Subtracting the second equation from the first we have,

(1.47) upper B left-parenthesis u minus u Subscript n Baseline comma v right-parenthesis identical-to upper B left-parenthesis e comma v right-parenthesis equals 0 for all v element-of upper S Superscript 0 Baseline left-parenthesis upper I right-parenthesis period

This equation is known as the Galerkin¹¹ orthogonality condition.

Theorem 1.4 If u Subscript n Baseline element-of upper S Superscript 0 Baseline left-parenthesis upper I right-parenthesis satisfies upper B left-parenthesis u Subscript n Baseline comma v right-parenthesis equals upper F left-parenthesis v right-parenthesis for all v element-of upper S Superscript 0 Baseline left-parenthesis upper I right-parenthesis then un minimizes the error u Subscript upper E upper X Baseline minus u Subscript n in energy norm where is the exact solution:

(1.48) vertical-bar vertical-bar vertical-bar vertical-bar minus minus u times times EXun Subscript upper E left-parenthesis upper I right-parenthesis Baseline equals min Underscript u element-of upper S overTilde Endscripts double-vertical-bar u Subscript upper E upper X Baseline minus u double-vertical-bar Subscript upper E left-parenthesis upper I right-parenthesis Baseline period

Proof: Let e equals u minus u Subscript n and let v be an arbitrary function in upper S Superscript 0 Baseline left-parenthesis upper I right-parenthesis . Then

vertical-bar vertical-bar vertical-bar vertical-bar plus plus ev Subscript upper E left-parenthesis upper I right-parenthesis Superscript 2 Baseline identical-to one half upper B left-parenthesis e plus v comma e plus v right-parenthesis equals one half upper B left-parenthesis e comma e right-parenthesis plus upper B left-parenthesis e comma v right-parenthesis plus one half upper B left-parenthesis v comma v right-parenthesis period

The first term on the right is double-vertical-bar e double-vertical-bar Subscript upper E left-parenthesis upper I right-parenthesis Superscript 2 , the second term is zero on account of Theorem 1.3, the third term is positive for any v not-equals 0 in upper S Superscript 0 Baseline left-parenthesis upper I right-parenthesis . Therefore double-vertical-bar e double-vertical-bar Subscript upper E left-parenthesis upper I right-parenthesis is minimum.

Theorem 1.4 states that the error depends on the exact solution of the problem u Subscript upper E upper X and the definition of the trial space ModifyingAbove upper S With tilde left-parenthesis upper I right-parenthesis .

The finite element method is a flexible and powerful method for constructing trial spaces. The basic algorithmic structure of the finite element method is outlined in the following sections.

1.3.1 The standard polynomial space

The standard polynomial space of degree p, denoted by script upper S Superscript p Baseline left-parenthesis upper I Subscript st Baseline right-parenthesis , is spanned by the monomials 1 comma xi comma xi squared comma ellipsis comma xi Superscript p defined on the standard element

(1.49) upper I Subscript st Baseline equals StartSet xi vertical-bar negative 1 less-than xi less-than 1 EndSet period

The choice of basis functions is guided by considerations

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